Question

You push a box out of a carpeted room and along a hallway with a waxed linoleum floor. While pushing the crate 2 mout of the room you exert a force of$\mathbf{34}\mathbf{}\mathit{N}$; while pushing italong the hallway you exert a force of$\mathbf{40}\mathbf{}\mathit{N}$. To slow it down you exert a force ofthrough a distance of$\mathbf{2}\mathbf{}\mathit{m}$, opposite to the motion. How much work do you do in all?

Answer

Short answer

$226N$

Step-by-step solution

Identification of the given data

The given data can be listed below as-

• The distance by which the crate moved in the first stage is$d_1=2m$,
• The force exerted on the crate in the first stage is$F_1=34N$,
• The distance by which the crate moved in the second stage is ,$d_2=6m$
• The force exerted on the crate in the second stage is,$F_2=13N$.
• The magnitude of the opposite force is,$F\text{'}=40N$$F\text{'}=40N$.

The distance moved by the crated when opposite force is applied is,$d\text{'}=2m$

Explanation of the work done

The work done for an object can be determined by taking the product of force applied on the object and the distance by which the object due to the application of force.

The equation of the work done is given as follows,

$W=Fd\cos \theta$

…(1)

Here, F is the force required to be exerted on the object, is the angle at which the force is applied and d is the distance by which the object moved from its position.

And, the total work done can be calculated by adding the separate work done.

Determination of the work done in first stage.

The expression for work done in the first stage is given as follows,

$W_1=F_1{d}_1\cos {\theta }_1$

Here, $F_1$isforce exerted on the crate in the first stage, $d_1$distance by which the crate moved in the first stage and${\theta }_1$ is the angle at which the force is applied.

For role="math" localid="1654107048846" $F_1=34N,{\theta }_1=0°and{d}_1=2m$

$$\begin{gathered} W_1=\left(34N\right)\left(2m\right)\cos 0° \\ =68Nm\times \frac{1J}{1Nm} \\ =68J \end{gathered}$$

Determination of the work done in second stage.

The expression for work done in the second stage is given as follows,

$W_2=F_2{d}_2\cos {\theta }_2$

Here, isforce exerted on the crate in the second stage, distance by which the crate moved in the second stage and is the angle at which the force is applied.

For $F_2=13N_1{\theta }_2=0°$, and $d_2=6m$.

$$\begin{gathered} W_2=\left(13N\right)\left(6m\right)\cos ° \\ =78Nm\times \frac{1J}{1Nm} \\ =78J \end{gathered}$$

Determination of the work done in application of force in opposite direction

The expression for work done when the force is applied in opposite direction is given as follows,

$W\text{'}=F\text{'}d\text{'}\cos \theta \text{'}$

Here, $F$isforce exerted on the crate in the opposite direction, $d$distance by which the crate moved when opposite force is applied and$\theta$ is the angle at which the force is applied.

For$F\text{'}=-40N,\theta =180°andd\text{'}=2m$ , and .

$$\begin{gathered} W\text{'}=\left(-40N\right)\left(2m\right)\cos 180° \\ =80Nm\times \frac{1J}{1Nm} \\ =80J \end{gathered}$$

Determination of the total work done

The expression for total work done is as follows,

$W=W_1=W_2=W\text{'}$

Substitute all the values in the above expression.